1. ## pole

Let $\displaystyle R(z)$ be a rational function:$\displaystyle R(z)=p(z)/q(z)$ where $\displaystyle p$ and $\displaystyle q$ are holomorphic polynomials. Let $\displaystyle f$ be holomorphic on $\displaystyle C$\ $\displaystyle \{P_1, P_2,....,P_k \}$ and suppose $\displaystyle f$ has a pole at each of the points $\displaystyle P_1, P_2,....,P_k$. Finally assume that $\displaystyle |f(z)| \leq |R(z)|$ for all $\displaystyle z$ at which these functions are defined. Prove that $\displaystyle f$ is a constant multiple of $\displaystyle R$.

2. Originally Posted by Kat-M
Let $\displaystyle R(z)$ be a rational function:$\displaystyle R(z)=p(z)/q(z)$ where $\displaystyle p$ and $\displaystyle q$ are holomorphic polynomials. Let $\displaystyle f$ be holomorphic on $\displaystyle C$\ $\displaystyle \{P_1, P_2,....,P_k \}$ and suppose $\displaystyle f$ has a pole at each of the points $\displaystyle P_1, P_2,....,P_k$. Finally assume that $\displaystyle |f(z)| \leq |R(z)|$ for all $\displaystyle z$ at which these functions are defined. Prove that $\displaystyle f$ is a constant multiple of $\displaystyle R$.

You meant, I presume, that $\displaystyle \{P_1,\ldots,P_k\}$ are the zeros of the denominator polynomial $\displaystyle q(z)$ , right? Well, we get that $\displaystyle q(z)f(z)$ is an entire function and

$\displaystyle |q(z)f(z)|\leq |p(z)|\Longrightarrow$ as no entire function can dominate another entire function unless one is a scalar multiple of the other (this follows at once from

Liouville's Theorem), we get what we want.

Tonio